Bunuel wrote:

What is the greatest value of n such that 30!/6^n is an integer?

A. 11

B. 12

C. 13

D. 14

E. 15

\(6^n\) can be written as \((2*3)^n\) = \(2^n *3^n\)

Now if we can figure how many 2 and 3 are present in 30 ! then we can have them removed. Cancelling(dividing) n number of 2 and 3 will remove n number of 6 (because 6 is a multiple of 2 and 3)

Therefore removing \(2^n\)and\(3^n\)will remove \(6^n\) and thus we can get the highest number of n that can be removed from 30!

2's in 30! =\(2^{26}\)

3's in 30!= \(3^{14}\)

so we can easily remove \(2^{14} *3^{14}\) from 30!

so \(6^{14}\) can be removed

therefore n=14

ANSWER IS D

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